The offset that brings the even orders back

Two boundaries removed, and one moved found a balanced differential pair’s mean and second harmonic at $10^{-17}$ — absent rather than small — and then unbalanced it by twenty millivolts and watched both come back to $10^{-2}$. That was a refusal, a check that the zeros belonged to the symmetry and not to the circuit. It said nothing about the shape of the return: whether a pair with a few hundred microvolts of offset has a second harmonic a few hundred times smaller than one with twenty millivolts, and where that proportionality stops.

What a pair cancels did not measure it either, and said so, offering an estimate instead: a millivolt of offset on a stage driven at twenty millivolts should restore “roughly a twentieth” of the single stage’s second harmonic. That is the number a designer needs — a transistor’s matching turned into its distortion — and it is worth having measured rather than estimated. The measurement below finds the estimate out by a factor of two and a half, and finds two boundaries the estimate had no way to see.

A closed form for the return #

An offset $V_{os}$ at the input of a pair moves the drive’s zero off the centre of the tanh. With the drive written $u$, its peak over $2V_T$, and the offset as $d = V_{os}/2V_T$, the output is $\tanh(d + u\sin\theta)$. Expanding about $d$, the fundamental comes from the slope there, $1 - \tanh^2 d$, and the second harmonic from the curvature, $-\tanh d\,(1 - \tanh^2 d)$, which a squared sinusoid turns into a cosine of twice the frequency with half its amplitude. Their ratio is

$$\frac{H_2}{H_1} = \frac{u}{2}\,\tanh\frac{V_{os}}{2V_T},$$

exact to second order in the drive and valid at any offset.

The solved waveforms follow it to three per cent at every drive below ten millivolts. The figure puts three curves on one axis. The pair’s third harmonic is the balanced pair’s own, $u^2/12$, and does not care about the offset: it reaches one per cent at 17.9 mV. The single transistor’s second harmonic is the reference, a quarter of its drive in thermal voltages. And the pair’s second harmonic runs parallel to it, a constant factor below, rising in proportion to the drive — until it does not. At a millivolt of offset it peaks at 0.70 per cent and turns down. The slider on the figure at the head of the page moves the offset from zero to ten millivolts.

What fraction of a transistor’s second harmonic comes back #

The constant factor between the dotted and solid curves is the result the estimate was reaching for, and it has a closed form of its own.

A single transistor driven at $a$ thermal voltages has a second harmonic of $a/4$ of its fundamental. The pair at the same drive has $u = a/2$ and so $(a/4)\tanh d$. The ratio is $\tanh(V_{os}/2V_T)$, with no drive in it, and it has a physical reading: it is the fraction of the tail current the offset steers to one side at rest. A perfectly balanced pair steers none and keeps none of the single transistor’s second harmonic; a pair so far off balance that one device carries the whole tail is a single transistor.

At a millivolt of offset that fraction is 1.93 per cent — a fifty-second, not a twentieth. The estimate was out by a factor of 2.6, in the optimistic direction for the single stage and the pessimistic one for the pair, so a millivolt of mismatch costs the pair less than it was said to: $-34.3$ dB against the single transistor, rather than $-26$. At 0.3 mV it is 0.58 per cent, $-44.7$ dB; at ten millivolts, 19.1 per cent. At twenty millivolts of drive the dots begin to leave the line, because at that drive the pair’s own higher-order terms have started to bend the tanh and the second-order expansion is no longer the whole curve.

Why the estimate said a twentieth #

The estimate’s reasoning is easy to reconstruct, and the error in it is instructive. A millivolt of offset on twenty millivolts of drive is a twentieth of the drive, and it is natural to guess that a twentieth of the asymmetry means a twentieth of the second harmonic. But the second harmonic does not come from the offset’s size against the drive. It comes from where the offset puts the operating point on the curve, and the curve’s scale is $2V_T$, not the drive. A millivolt is a fifty-second of $2V_T$ at room temperature, and that is the fraction of the tail steered aside and the fraction of a single transistor’s second harmonic that returns, at any drive at which the expansion holds.

So the right comparison for an offset is always with the thermal voltage. A pair whose offset is small against $2V_T$ is nearly balanced however small the signal, and one whose offset is comparable to $2V_T$ is nearly a single transistor however large the signal. The drive enters only through the single transistor’s own second harmonic, which both share — the distortion a linear model cannot have measured that as a quarter of the drive in thermal voltages, and the pair keeps $\tanh(V_{os}/2V_T)$ of it.

First order, then a ceiling #

The closed form says the second harmonic is proportional to the offset while $\tanh d \approx d$, and says exactly where that stops.

At ten millivolts of drive the solved second harmonic follows $(u/2)\tanh d$ to three per cent over four decades of offset, from ten microvolts to two hundred millivolts. It is proportional to the offset — the dashed line — to within one per cent up to 9.0 mV, which is where $\tanh d$ is one per cent below $d$, about a third of a thermal voltage. Above that the proportionality bends and the second harmonic saturates at $u/2$, 9.7 per cent here: at that point one device carries the tail and the pair has become a single transistor.

So for every offset a real monolithic pair has — tens of microvolts to a few millivolts — the second harmonic is strictly first order in the offset. A matching specification translates into a distortion specification by a factor, $u/4V_T$ per volt of offset, and nothing else. That is the gentlest failure a symmetry can have: it degrades in proportion to how badly it is broken, which how small is small signal would recognise as the same kind of statement about a different expansion — a first-order term, valid until a fixed fraction of the thermal voltage.

The mean comes back with it #

The second harmonic is not the only even-order quantity an offset restores. The same curvature that turns a squared sinusoid into a second harmonic turns it into a constant as well, so the pair’s mean output — the operating point it actually sits at while driven — moves with the square of the drive. To second order the shift is $-\tanh d\,(1 - \tanh^2 d)\,u^2/2$ of the tail: zero for a balanced pair, where two boundaries removed, and one moved found it at $10^{-17}$, and first order in the offset for a small one.

This is the pair’s version of the boundary the point the device is never at measured for a single transistor, whose mean current rises with the square of the drive. A single transistor’s shift is a quarter of the squared drive in thermal voltages, as a fraction of its current; the imbalanced pair’s, counted in the same current per device, is smaller by exactly $\tanh d$, so a millivolt of offset leaves the pair’s operating point about fifty times steadier than a single transistor’s under the same signal. Of the single transistor’s three boundaries, an offset restores the two a pair had removed, both scaled by the same fraction of steered tail, and moves neither of them far.

The boundary that does not always exist #

The single transistor’s second harmonic reaches one per cent at 0.04 thermal voltages of drive, about a millivolt, and grows for ever after. The pair’s second harmonic, being $(u/2)\tanh d$ at small drive, ought to reach one per cent at a drive of $2V_T \cdot 0.02/\tanh d$ — 53.5 mV at a millivolt of offset. It does not.

The first-order form assumes the drive is small against $2V_T$, and 53.5 mV is twice that. At such drives the pair is beginning to switch: its output is flattening towards a square wave, and a square wave driven off-centre has small even harmonics, not large ones. The second harmonic therefore rises with the drive, peaks near 76 mV, and falls again. At a millivolt of offset the peak is 0.70 per cent; at 0.3 mV, 0.21 per cent. Below an offset of 1.44 mV the peak never reaches one per cent at all, and the one-per-cent second-harmonic boundary does not exist.

That is a third boundary the earlier refusal could not see, and it is the right way round for a designer. A pair matched to better than a millivolt and a half cannot be driven into one per cent of second harmonic by any signal; it will saturate first. The distortion it does produce at large drive is odd: the third harmonic, which reaches one per cent at 17.9 mV whatever the offset.

The two cross where the second harmonic’s boundary falls to the third’s. The solved crossing is at 3.14 mV of offset. Below it a pair is limited by its odd orders, exactly as a balanced pair is, and its matching is invisible in its distortion budget. Above it the second harmonic reaches one per cent first, at a drive inversely proportional to the offset — 18.8 mV at three millivolts of offset, 5.4 mV at ten — and the pair’s distortion is its matching’s.

A matching specification, read as a distortion one #

The closed form inverts: for a stated second harmonic $h$ at a stated drive, the largest offset allowed is $2V_T\,\operatorname{atanh}(2h/u)$.

Each curve falls as one over the drive, since the second harmonic is the offset times the drive. At ten millivolts of drive, $-40$ dB of second harmonic allows 5.37 mV of offset, $-60$ dB 0.535 mV, and $-80$ dB 53.5 µV. The last is the matching of a good monolithic pair with its offset trimmed; the first is an untrimmed discrete pair. The solved pair, set to the offset each curve allows, returns the stated level to a third of a decibel at the three points checked.

The practical reading is that a pair’s offset is a fixed property of its matching and its distortion is that offset times the drive. A pair matched well enough for its second harmonic at one signal level is ten times worse at ten times the level, until it saturates. And the budget is a straight trade: every decibel of second harmonic the application can tolerate at a given drive is a decibel of offset the pair is allowed.

A discrete pair and a monolithic one #

Put numbers on the two pairs a designer actually meets. An untrimmed pair of discrete transistors might have five millivolts of offset. That steers 9.6 per cent of its tail aside and keeps the same fraction of a single transistor’s second harmonic, $-20.3$ dB against it. Its second harmonic reaches one per cent at 10.7 mV of drive — before its third harmonic does at 17.9 — so its distortion is its matching’s, and a designer who chose a pair to be rid of the even orders has been rid of nine-tenths of them.

A monolithic pair on one die, with three hundred microvolts of offset, keeps 0.58 per cent of the single transistor’s second harmonic, $-44.7$ dB against it. It never reaches one per cent of second harmonic at any drive; it saturates first, at a peak of 0.21 per cent. Its distortion budget is its third harmonic’s, and its matching is invisible in it.

The two are called by the same name and drawn with the same symbol. The rejection the parts have made the same observation about an instrumentation amplifier’s common-mode rejection: the topology sets the form of the answer and the matching sets its size, and for a pair’s even orders the size spans twenty-four decibels between a discrete and a monolithic part.

What a designer should take #

A pair’s second harmonic is $(u/2)\tanh(V_{os}/2V_T)$ of its fundamental: a single transistor’s second harmonic times the fraction of its tail the offset steers aside, first order in the offset for any offset below about 9 mV. Use it to turn a matching specification into a distortion specification directly. A millivolt of offset keeps 1.9 per cent of a single transistor’s second harmonic, a fifty-second of it.

Two boundaries come with it. Below about 1.4 mV of offset the second harmonic never reaches one per cent, because the pair saturates first. Above about 3.1 mV it reaches one per cent before the balanced pair’s third harmonic does, at 17.9 mV, and the offset rather than the transfer curve limits the stage’s distortion. Between the two, both matter, and the drive decides which binds.

And since the offset of a bipolar pair is the thermal voltage times the log of its saturation-current ratio, as what matching does about temperature measured, $V_{os}/2V_T = \tfrac12 \ln m$ contains no temperature at all. The second harmonic a mismatch restores is therefore the same at every temperature for a fixed drive in thermal voltages: the offset drifts, and its distortion does not.

How the numbers were obtained #

The pair’s output is $\tanh(u\sin\theta + d)$ evaluated at 1,024 points of a period, and its fundamental and harmonics are read from a discrete transform; the single transistor’s harmonics are the modified Bessel functions of its drive, which the earlier essays checked against their own transforms. Each boundary is found by scanning drive geometrically from half a millivolt in steps of four per cent and bisecting in log drive between the last step below and the first at or above one per cent, forty iterations. The offset below which the boundary does not exist is bisected on the largest second harmonic over that scan, and the crossing with the third harmonic on the located boundary. The temperature is 27 °C throughout.

What it leaves out #

Degeneration. Emitter resistors reduce the second harmonic an offset restores, since they flatten the curvature the offset exposes, and symmetry and a resistor together found them leaving the balanced pair’s even orders at zero. The offset’s version with the resistors in has a closed form of its own, not drawn here.

Mismatch that is not an offset. A difference in the two devices’ saturation currents is an input offset exactly. A difference in their emitter resistors or their current gains is not: it changes the two halves’ slopes as well as their centre, and the second harmonic it restores grows with the drive differently.

The tail. An ideal tail is assumed, so the two collector currents always sum to the same total and either collector alone carries only odd harmonics of a balanced pair. A real tail has a finite impedance, and a pair driven on one input with the other held moves its tail node with the signal; the tail current then varies at twice the signal frequency, and that is a second harmonic the matching does not control.

Still open: the degenerated offset, the single-ended output, and the mismatched gains #

The offset with the emitter resistors in. With degeneration the second harmonic an offset restores should fall faster than the fundamental, by some power of $1 + g_m R_e$. Measuring it would say whether degenerating a pair relaxes its matching specification, and by how much for each decibel of gain given up.

A single-ended drive with a real tail. Driving one input and holding the other moves the tail node, and through the tail’s finite impedance the tail current follows the square of the drive. That puts a second harmonic into a perfectly matched pair; its size against the offset’s contribution, as a function of the tail impedance, decides whether matching or the tail limits a single-ended stage.

Current gains that differ. A difference in β between the two devices makes their base currents differ and so loads a source unequally; with a source resistance it becomes an offset that depends on the signal. Whether that restores even orders faster or slower than a fixed offset of the same size is a solve with the source resistance in it.